Download MATH 522–01 Problem Set #1 solutions 1. Let U be a nonempty set

MATH 522–01
Problem Set #1 solutions
1. Let U be a nonempty set and let R be the set of all subsets of U (i.e. the power set of U ). For the
two given proposed definitions of “addition” and “multiplication”, determine whether R is a ring or
not; if it is not a ring, explain why, and if it is a ring, identify its identity elements.
(a) A + B = A ∪ B and A · B = A ∩ B.
This is not a ring, because there is no inverse under addition. To determine an identity, we note
that A + 0 = A ∪ 0 = A is only universally true if 0 = ∅; however, then since for nonempty A,
there is no set (−A) such that A + (−A) = A ∪ (−A) = ∅.
(b) A + B = (A ∪ B) − (A ∩ B) (also known as the “symmetric difference”) and A · B = A ∩ B.
This is in fact a ring: closure is easily demonstrated, and associativity is a bit tedious but
straightforward (A + B + C consists of those elements lying in exactly 1 or 3 of the sets A, B,
and C). The distributeive law may be derived from noting that [(A ∪ B) − (A ∩ B)] ∩ C =
[(A ∩ C) ∪ (B ∩ C)] − (A ∩ B ∩ C) The additive identity is ∅, and every element of R is its own
additive inverse. There is also a multiplicative identity, U itself.
2. Prove that for a commutative ring R and x ∈ R, x is a unit of R if and only if x divides every
element of R.
First, let us note that the statement is vacuously true if R lacks a multiplicative identity, as neither
of the conditions can be met. Clearly, x can be a unit only if xy = 1 for some x, y ∈ R, which
would require that 1 ∈ R. Similarly, if x divides every element of R, then in particular x divides
x, so x = kx for some k ∈ R; now, for any r ∈ R, there is an ℓ ∈ R such that r = ℓx, and so
r = ℓx = ℓkx = k(ℓx) = kr, making k a multiplicative identity. Henceforth, we may thus assume
that R has a multiplicative identity.
Let us suppose that x is a unit, so that there is a y ∈ R such that xy = 1. Then for any r ∈ R,
(xy)r = 1r = r, so r = x(yr). Thus x divides r.
Conversely, if we suppose that x divides every element of R, then, in particular, x divides 1, so there
is a k ∈ R such that xk = 1; since k is definitionally the inverse of x, x is a unit.
3. Prove the following two statements:
(a) For any ring R and a (not necessarily finite) collection R of subrings of R, their intersection
∩
S
S∈R
is also a subring of R. (Hint: you may find it easier to warm up by addressing just an intersection
of two subrings of R)
∩
Let T = S∈R S for brevity. Then if a, b ∈ T , it follows that a, b ∈ S for every single ring
S ∈ R. Since each S is a ring, ab and a + b are in S for every S ∈ R, and thus ab and a + b are
in T , demonstrating closure of T under addition and multiplication.
Likewise, since each S is a ring, −a ∈ S for every S ∈ R, and so −a ∈ T , demonstrating closure
under additive inversion.
Finally, since every S is a ring, 0 ∈ S and thus 0 ∈ T .
(b) For a ring R and a subset S of R, there is a unique smallest subring ⟨S⟩ of R which contains
all the elements of S (hint: choose an appropriate collection R for the above statement).
Let R
∩ contain every ring which is a superset of S and a subring of R. By the previous part,
T = X∈R X is a subring of R, and for every ring X which is a superset of S and a subring of
R, X ⊆ T , we may assert that T , as it contains S and is contained in every ring containing S,
is the smallest such ring.
Page 1 of 2
due Thursday, January 22, 2015
MATH 522–01
Problem Set #1 solutions
4. Address the two following questions about subrings:
√ √
(a) Let Q[ 2, 3] denote
√ the smallest subring of R which
√ contains√every√rational number, the
irrational number 2, and the irrational number 3. Let Q[ 2 + 3] denote the
√ small√
est subring of R which contains every rational number and the irrational number 2 + 3
(note
√ that
√ the previous
√
√ question guarantees that these structures are well-defined). Prove that
Q[ 2, 3] = Q[ 2 + 3].
√
√
√
√
√ √
It √
suffices
to
show
that
2
and
3
are
elements
of
Q[
2
+
3]
to
prove
that
Q[
2,
√
√
√
√ √
√
√
√ 3]
√⊆
Q[ 2 + 3]; likewise, if we show that 2 + 3 ∈ Q[ 2, 3], then Q[ 2 + 3] ⊆ Q[ 2, 3],
collectively showing the desired equality.
√
√ √
√
√ √
The
second
of
these
is
trivial:
2
∈
Q[
2,
3]
and
3
∈
Q[
2, 3], so by closure of addition,
√
√
√ √
2 + 3 ∈ Q[ 2, 3].
√
√
√
√
The first membership, however, is more difficult. Since
(
2
+
3)
∈
Q[
2
+
√
√
√ 3], we may use
closure of multiplication to find that its square 5 + 2 6 is also in Q[ 2 + 3]. −5 and 12 are
√
√
√
6
Q[
2
+
rational, and can be added
and
multiplied
in
turn
to
the
above
to
find
that
∈
√
√
√
√
√
√
√3].
Finally, we multiply 6 by ( 2 +√ 3)√once more to find that√2 3 + 3 √2 is in Q[ 2 + 3],
and
twice and thrice ( 2 + 3) demonstrates that 2 and − 3 respectively are in
√subtracting
√
Q[ 2 + 3].
√ √
√
√
(b) Let Z[ 2, 3] and Z[ 2 + 3] be defined as in the previous √
part√except with
number”
√ “rational
√
replaced by “integer” in the definition. Demonstrate that Z[ 2, 3] ̸= Z[ 2 + 3].
√
√
√ √
The same argument as above can be used to show that Z[ 2 + 3] ⊆ Z[ 2, 3]; however, our
argument for the reverse containment will not work as designed because we cannot multiply
by 21 . It is, however, easy to follow the above argument without the division to guarantee that
√
√
√
√
√
2 2, 2 3, and 2 6 are all elements of Z[ 2 + 3]. However, half of each of these values is not
in the ring: note that the set
√
√
√
√
{a + 2b 2 + 2c 6 + d( 2 + 3) : a, b, c, d ∈ Z}
√
√
√
√
is a ring containing Z and 2 + 3 but not 2 or 3.
5. Prove that if the additive group of a ring R is cyclic, then R is commutative.
Let g be a generator of the additive group. Thus, for any a, b ∈ R, either a or −a may be written
as g + g + · · · + g for some nonnegative integer k; likewise b or −b may be written as g + g + · · · + g
|
{z
}
|
{z
}
k times
ℓ times
for some non-negative integer ℓ. Using the distributive law, the product ab can be written as
± g + g + · · · + g ; however, likewise ba could be written as ± g + g + · · · + g ; since k and ℓ are ordi|
{z
}
|
{z
}
kℓ times
ℓk times
nary integers, the number of addends is the same in both expressions, so ab = ba.
Page 2 of 2
due Thursday, January 22, 2015